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A polynomial \(\varphi(s) = a_{n}s^{n} + a_{n-1}s^{n-1} + \cdots + a_{1}s + a_{0}\) of degree \(n \gt 3\) with constant real coefficients \(a_{n}, a_{n-1}, \ldots a_{0}\) has triple roots at \(s = -\sigma\). Which one of the following conditions must be satisfied?

  • A.\(\varphi(s) = 0\) at all the three values of \(s\) satisfying \(s^{3} + \sigma^{3} = 0\)
  • B.\(\varphi(s) = 0\), \(\dfrac{d\varphi(s)}{ds} = 0\), and \(\dfrac{d^{2}\varphi(s)}{ds^{2}} = 0\) at \(s = -\sigma\)
  • C.\(\varphi(s) = 0\), \(\dfrac{d^{2}\varphi(s)}{ds^{2}} = 0\), and \(\dfrac{d^{4}\varphi(s)}{ds^{4}} = 0\) at \(s = -\sigma\)
  • D.\(\varphi(s) = 0\), and \(\dfrac{d^{3}\varphi(s)}{ds^{3}} = 0\) at \(s = -\sigma\)

Answer: B

The condition is \(\varphi(s) = 0\), \(d\varphi/ds = 0\) and \(d^{2}\varphi/ds^{2} = 0\) at \(s = -\sigma\).

A triple root at \(s = -\sigma\) means \(\varphi(s) = (s + \sigma)^{3}q(s)\), where \(q(-\sigma) \ne 0\). Differentiating, the factor \((s + \sigma)\) survives in \(\varphi\), \(\varphi'\) and \(\varphi''\), so all three vanish at \(-\sigma\); in \(\varphi'''\) it does not, so the third derivative is in general non-zero.

The option on \(s^{3} + \sigma^{3} = 0\) is wrong: that equation has \(-\sigma\) and two complex roots, and a repeated real root says nothing about those complex points. The option with the second and fourth derivatives leaves out the first derivative, which must vanish, and adds a fourth, which need not. The option with the third derivative is wrong because a root of multiplicity exactly three is the case where the third derivative does not vanish.

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